By Wu Peng, Senior Process Instrumentation Engineer · Last reviewed July 26, 2026
Home › Tools › 4-20 mA to voltage calculator
This calculator turns a 4-20 mA loop signal into volts and back. Four modes: current to voltage across a burden resistor, the resistor needed for a target range, a measured voltage back to current and process value, and a loop compliance check. The core is V = I x R, with the loop’s own limits checked around it.
One result surprises people. A resistor cannot produce a true 0 V from 4 mA, so a 500 ohm burden gives 2-10 V rather than 0-10 V. The calculator says so, and gives the gain and offset an active converter needs instead. The reasoning is in the guide on converting 4-20 mA to 0-10 V, 1-5 V and 0-5 V.
Calculator
4-20 mA and voltage converter
Results update as you type. Signal status uses the NAMUR NE43 bands: valid 3.8 to 20.5 mA, downscale fault at 3.6 mA and below, upscale fault at 21.0 mA and above.
How the math works
A 4-20 mA loop is a current signal, so every voltage input reads it indirectly: the current runs through a known resistance and the input measures the drop across it.
V = I x R and back the other way I = V / R
ANSI/ISA-50.1-1982 (R1992) names 250 ohm as the standard load resistor and permits 50 to 1000 ohm. That is why 1-5 V is the classic voltage form of the signal. The quick check every technician knows: 250 ohm gives 1 V at 4 mA, 3 V at 12 mA and 5 V at 20 mA. The calculator reproduces those exactly.

Worked example
A 0-300 psi transmitter feeds a data logger through a 250 ohm resistor, and a meter across the resistor reads 3.42 V. Mode three answers what that means.
- Loop current: 3.42 V / 250 ohm = 13.68 mA
- Percent of span: (13.68 − 4) / 16 = 0.605, so 60.5 percent
- Process value: 0 + 0.605 x 300 = 181.5 psi
- Signal status: inside 3.8 to 20.5 mA, so valid
Now run the same loop the other way. Mode two, with a target of 0 V at 4 mA and 10 V at 20 mA, returns 500 ohm for full scale. It also warns that a resistor gives 2.000 V at 4 mA, and prints the active-converter setting of 0.625 V/mA with a −2.5 V offset.
Mode four then checks whether a 500 ohm burden is legal on that loop. On a 24 V supply with a 12 V transmitter and 30 ohm of cable, the budget is 600 ohm, the loop uses 530 ohm, and 70 ohm of headroom is left. Move the supply down to 12 V and the same burden fails outright.
Application example
Aviation fuel handling, Poland. A customer measuring Jet A1 fuel level specified a 360 mm capacitive probe, flanged, with a 0-5 V output, because the receiving unit had a voltage input. That requirement was fixed at the enquiry stage, so no burden resistor entered the design. The probe was specified with the voltage output built in, and the open questions moved to pressure rating and flange specification.
Choosing the resistor
Three checks decide the value, and the calculator covers all three.
| Check | What to look at |
|---|---|
| Card range | 249 ohm for a 1-5 V or 0-5 V input, 499 ohm for 0-10 V. Standard 1 percent values sit 0.4 and 0.2 percent below nominal full scale, so put the real value into the scaling |
| Loop compliance | Total loop resistance must stay under (supply − transmitter minimum) / 0.020 A, including cable, barriers and indicators |
| Fault levels | An upscale fault at 21 mA reaches 10.50 V across 500 ohm and clips on a 0-10 V input; across 250 ohm it is 5.25 V and stays visible on a 0-10 V card |
Specify metal film, 0.1 percent tolerance, low temperature coefficient, 0.25 W or more, mounted at the receiving device.
Resolution on the card
A voltage that only fills part of the card range costs resolution, and the loss is easy to miss because nothing looks wrong. Only the fraction of the input range the signal occupies produces counts.
| Signal into the card | Share of range | Counts, 12-bit | Steps on 0-300 psi |
|---|---|---|---|
| 1-5 V into 0-5 V, or 2-10 V into 0-10 V | 80 percent | 3,276 of 4,095 | 0.09 psi |
| 1-5 V into 0-10 V | 40 percent | 1,638 of 4,095 | 0.18 psi |
| 1-5 V into ±10 V | 20 percent | 819 of 4,095 | 0.37 psi |
Counts assume an ideal converter. A 16-bit card multiplies every figure by 16, which usually settles the question.
So the burden that matches your card range is worth more than the cheapest value in stock. Run mode one with your resistor and read the 4 mA and 20 mA endpoints, then compare that span against the card range before you order the part.
Related guides and tools
The full method, including input impedance loading, fault levels in volts and wiring practice, is in convert 4-20 mA to 0-10 V, 1-5 V and 0-5 V. For loop design and fault currents, read the 4-20 mA current loop guide. To go straight from current to engineering units, use the 4-20 mA calculator. Terminal-by-terminal layouts are in the pressure transducer wiring guide. If the receiving system takes voltage natively, the SI-390 industrial pressure transducer and the SI-303 low pressure transducer are ordered with 1-5 V, 0-5 V or 0-10 V outputs.
FAQ
Can you convert 4/20mA to 0 10V?
Yes, but not with a resistor alone. A 500 ohm resistor produces 2-10 V because 4 mA has to drop 2 V across it. To reach a true 0-10 V, either rescale 2-10 V as 0-100 percent in the controller or use an active converter set to 0.625 V/mA with a −2.5 V offset.
How to convert mA to volts?
Multiply the current in amps by the resistance in ohms: 12 mA through 250 ohm is 0.012 x 250 = 3.000 V. Mode one does this and also shows the 4 mA and 20 mA endpoints, the NAMUR status of the current and the power dissipated in the resistor.
How to calculate 4 to 20 mA?
Current from a process value is I = 4 + 16 x (PV − PVmin) / (PVmax − PVmin). The reverse is PV = PVmin + (I − 4) / 16 x (PVmax − PVmin). Mode three applies the reverse form to a measured voltage, so a meter reading becomes psi, degrees or cubic meters per hour in one step.
How to calculate shunt resistor value?
Divide the voltage you want at full scale by 0.020 A: a 5 V full scale needs 250 ohm, a 10 V full scale needs 500 ohm. Then check the loop can afford that resistance and pick a standard 1 percent part, typically 249 ohm or 499 ohm.
Request a quote
Tell us what the receiving system accepts and how far the signal has to travel, and we configure the instrument output around it rather than leaving a resistor to correct it at the terminals. Reach our application engineers or use the form below.
Written and technically reviewed by Wu Peng and the Instranova engineering team.